[@integralsbot](https://twitter.com/integralsbot)さんがツイートした[こちらの定理](https://twitter.com/integralsbot/status/1368225055826583552)の解説です．

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以下$2$つの等式が成り立ちます．ただし$\alpha\in\R$，$\beta\in\R$とし，$-\pi<\alpha<\pi$を満たすとします．
- $\d\i{x}{-\infty}{\infty}{\frac{\sinh\alpha x}{\sinh\pi x}\cos\beta x}=\frac{\sin\alpha}{\cos\alpha+\cosh\beta}$
- $\d\i{x}{-\infty}{\infty}{\frac{\cosh\alpha x}{\sinh\pi x}\sin\beta x}=\frac{\sinh\beta}{\cos\alpha+\cosh\beta}$
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&&&prf 
<details><summary>解説</summary>
\begin{align*}
&\i{x}{0}{\infty}{\frac{\sinh\r{\alpha+i\beta}x}{\sinh\pi x}}\\
=&\i{x}{0}{\infty}{\frac{\frac{e^{\r{\alpha+i\beta}x}-e^{-\r{\alpha+i\beta}x}}{2}}{\frac{e^{\pi x}-e^{-\pi x}}{2}}}\\
=&\i{x}{0}{\infty}{\frac{e^{\r{\alpha-\pi+i\beta}x}-e^{\r{-\alpha-\pi-i\beta}x}}{1-e^{-2\pi x}}}\\
=&\i{x}{0}{\infty}{\r{e^{\r{\alpha-\pi+i\beta}x}-e^{\r{-\alpha-\pi-i\beta}x}}\su{n}{0}{\infty}{e^{-2\pi nx}}}\\
=&\i{x}{0}{\infty}{\su{n}{0}{\infty}{\r{e^{\r{\alpha-\r{2n+1}\pi+i\beta}x}-e^{\r{-\alpha-\r{2n+1}\pi-i\beta}x}}}}\\
=&\s{\su{n}{0}{\infty}{\r{\frac{e^{\r{\alpha-\r{2n+1}\pi+i\beta}x}}{\alpha-\r{2n+1}\pi+i\beta}-\frac{e^{\r{-\alpha-\r{2n+1}\pi-i\beta}x}}{-\alpha-\r{2n+1}\pi-i\beta}}}}_{x=0}^\infty\\
=&-\su{n}{0}{\infty}{\r{\frac{1}{\alpha+i\beta-\r{2n+1}\pi}+\frac{1}{\alpha+i\beta+\r{2n+1}\pi}}}\\
=&-\su{n}{0}{\infty}{\frac{2\r{\alpha+i\beta}}{\r{\alpha+i\beta}^2-\r{\r{2n+1}\pi}^2}}\\
=&\frac{1}{2\pi}\r{-\su{n}{0}{\infty}{\frac{2\r{\frac{\alpha+i\beta}{2\pi}}}{\r{\frac{\alpha+i\beta}{2\pi}}^2-\r{n+\half}^2}}}\\
=&\frac{1}{2\pi}\pi\tan\pi\frac{\alpha+i\beta}{2\pi}\\
=&\half\tan\frac{\alpha+i\beta}{2}\\
=&\half\frac{2\sin\frac{\alpha+i\beta}{2}\cos\frac{\alpha-i\beta}{2}}{2\cos\frac{\alpha+i\beta}{2}\cos\frac{\alpha-i\beta}{2}}\\
=&\half\frac{\sin\alpha+\sin i\beta}{\cos\alpha+\cos i\beta}\\
=&\half\frac{\sin\alpha+i\sinh\beta}{\cos\alpha+\cosh\beta}
\end{align*}
より，
\begin{align*}
&\i{x}{-\infty}{\infty}{\frac{\sinh\alpha x}{\sinh\pi x}\cos\beta x}\\
=&\Re\i{x}{-\infty}{\infty}{\frac{\sinh\alpha x\cos\beta x+i\cosh\alpha x\sin\beta x}{\sinh\pi x}}\\
=&\Re\i{x}{-\infty}{\infty}{\frac{\sinh\alpha x\cosh i\beta x+\cosh\alpha x\sinh i\beta x}{\sinh\pi x}}\\
=&\Re\i{x}{-\infty}{\infty}{\frac{\sinh\r{\alpha+i\beta}x}{\sinh\pi x}}\\
=&\Re2\i{x}{0}{\infty}{\frac{\sinh\r{\alpha+i\beta}x}{\sinh\pi x}}\\
=&\Re2\half\frac{\sin\alpha+i\sinh\beta}{\cos\alpha+\cosh\beta}\\
=&\frac{\sin\alpha}{\cos\alpha+\cosh\beta}
\end{align*}
及び
\begin{align*}
&\i{x}{-\infty}{\infty}{\frac{\cosh\alpha x}{\sinh\pi x}\sin\beta x}\\
=&\Im\i{x}{-\infty}{\infty}{\frac{\sinh\alpha x\cos\beta x+i\cosh\alpha x\sin\beta x}{\sinh\pi x}}\\
=&\Im\i{x}{-\infty}{\infty}{\frac{\sinh\alpha x\cosh i\beta x+\cosh\alpha x\sinh i\beta x}{\sinh\pi x}}\\
=&\Im\i{x}{-\infty}{\infty}{\frac{\sinh\r{\alpha+i\beta}x}{\sinh\pi x}}\\
=&\Im2\i{x}{0}{\infty}{\frac{\sinh\r{\alpha+i\beta}x}{\sinh\pi x}}\\
=&\Im2\half\frac{\sin\alpha+i\sinh\beta}{\cos\alpha+\cosh\beta}\\
=&\frac{\sinh\beta}{\cos\alpha+\cosh\beta}
\end{align*}
なので，$\d\i{x}{-\infty}{\infty}{\frac{\sinh\alpha x}{\sinh\pi x}\cos\beta x}=\frac{\sin\alpha}{\cos\alpha+\cosh\beta}$及び$\d\i{x}{-\infty}{\infty}{\frac{\cosh\alpha x}{\sinh\pi x}\sin\beta x}=\frac{\sinh\beta}{\cos\alpha+\cosh\beta}$です．$\blacksquare$
</details>
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